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1. đề bạn ghi rõ lại giúp mình đc ko r mình giải lại cho
2. Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x^2}{2.3^2}=\dfrac{y^2}{5^2}=\dfrac{2x^2-y^2}{18-25}=\dfrac{-28}{-7}=4\)
\(\dfrac{x}{3}=4\Rightarrow x=12\)
\(\dfrac{y}{5}=4\Rightarrow y=20\)
Vậy x=12 và y=20
Bài 1:
a, \(\dfrac{x+5}{x}=\dfrac{4}{3}\)
\(\Rightarrow3x+15=4x\\ \Rightarrow4x-3x=15\\ \Rightarrow x=15\)
b, \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
\(\Rightarrow\left(x-20\right).\left(x+70\right)=\left(x+40\right)\left(x-10\right)\)
\(\Rightarrow x^2+70x-20x-1400=x^2-10x+40x-400\)
\(\Rightarrow x^2-x^2+70x-20x+10x-40x=-400+1400\)
\(\Rightarrow20x=1000\Rightarrow x=50\)
c, \(4^x=\dfrac{1.2.3.....31}{4.6.8.....64}\)
\(\Rightarrow4^x=\dfrac{1}{2.2.2.2.....2.2.64}\) (có 30 số 2)
\(\Rightarrow4^x=\dfrac{1}{2^{30}.4^3}\Rightarrow4^x=\dfrac{1}{4^{15}.4^3}\)
\(\Rightarrow4^x=\dfrac{1}{4^{18}}\)
\(\Rightarrow4^x=4^{-18}\)
Vì \(4\ne-1;4\ne0;4\ne1\) nên \(x=-18\)
Chúc bạn học tốt!!!
a , \(\dfrac{x+5}{x}=\dfrac{4}{3}\Leftrightarrow3\left(x+5\right)=4x\)
<=> 3x+15=4x
<=> x= 15
b , \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
<=> \(\dfrac{x-10}{x-10}-\dfrac{10}{x-10}=\dfrac{x+70}{x+70}-\dfrac{30}{x+70}\)
<=> \(1-\dfrac{10}{x-10}=1-\dfrac{30}{x+70}\)
<=> \(\dfrac{10}{x-10}=\dfrac{30}{x+70}\Leftrightarrow\dfrac{1}{x-10}=\dfrac{3}{x+70}\)
<=> (x+70)=3(x-10)
<=> x+70 = 3x-30
<=> 100=2x
<=> x= 50
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi
Kêu người ta giúp mà ói vào mặt người ta vậy à?
b, \(\dfrac{x^3+y^3}{6}\) = \(\dfrac{x^3-2y^3}{4}\)và x6.y6=64
=>(x3+y3 ).4=(x3-2y3).6
=>4x3+4y3=6x3-12y3
=> 4y3 + 12y3= 6x3-4x3
=> 15y3=2x3
Làm được thế này thoy
(lm đc hết thì càng tốt ^-^)
1. a) (x-2)2 =1
=> x - 2 = \(\pm\sqrt{1}\)
=> x - 2 = 1 hoặc -1
=> x = 3 hoặc 1
b) 2x - 1= -8
=> 2x = -7
=>x = \(\dfrac{-7}{2}\)
c)thiếu đề
d) (x-1)x+2 = (x-1)x+4
(x-1)x+2 = (x-1)x+2+2
(x-1)x+2 = (x-1)x+2. (x-1)2
(x-1)x+2 - (x-1)x+2. (x-1)2 = 0
=> (x-1)x+2. [1 - (x-1)2] = 0
\(\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2a) \(\dfrac{45^{10}.5^{10}}{75^{10}}\) = \(\dfrac{\left(3.3.5\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\dfrac{3^{10}.3^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(3^{10}\)
b) \(\dfrac{2^{15}.9^4}{6^6.8^3}\)=\(\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)=\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)=\(3^2\)
Câu 2:
a) \(\sqrt{x}=5\)
\(\Leftrightarrow x=25\)
b) \(2\sqrt{x}=\sqrt{12}\)
\(\Leftrightarrow2\sqrt{x}=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}=\sqrt{3}\)
\(\Leftrightarrow x=3\)
c) \(x^2=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\)
d) \(-3\sqrt{x}=-\sqrt{18}\)
\(\Leftrightarrow-3\sqrt{x}=3\sqrt{2}\)
\(\Leftrightarrow\sqrt{x}=\sqrt{2}\)
\(\Leftrightarrow x=2\)
e) \(x^2-1=7\)
\(\Leftrightarrow x^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\end{matrix}\right.\)
f) \(3\sqrt{x^2}=\sqrt{9}\)
\(\Leftrightarrow3\cdot\left|x\right|=3\)
\(\Leftrightarrow\left|x\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a, Ta có: \(\left(xyz\right)^2=\dfrac{2}{7}.\dfrac{3}{2}.\dfrac{3}{7}\)\(=\dfrac{9}{49}\)
\(\Rightarrow xyz=\sqrt{\dfrac{9}{49}}=\dfrac{3}{7}.\)
\(\Rightarrow z=\dfrac{xyz}{xy}=\dfrac{3}{7}:\dfrac{2}{7}=1,5.\)
\(\Rightarrow y=1;x=\dfrac{2}{7}\).
b, Tương tự.
a) Ta có:
\(\left|x-2017\right|\ge0\) với \(\forall x\)
\(\left|y-2018\right|\ge0\) với \(\forall x\)
\(\Rightarrow\left|x-2017\right|+\left|y-2018\right|\ge0\) với \(\forall x\)
\(\Rightarrow\) Không có giá trị của x; y thỏa mãn yêu cầu
Vậy \(x;y\in\varnothing\)
b) Ta có:
\(3.\left|x-y\right|^5\ge0\)
\(10.\left|y+\dfrac{2}{3}\right|^7\ge0\)
\(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\ge0\left(1\right)\)
Theo bài ra ta có: \(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\le0\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7=0\)
\(\Rightarrow\left\{{}\begin{matrix}3.\left|x-y\right|^5=0\\10.\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|^5=0\\\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x-y=0\\y+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=y\\y=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{-2}{3}\end{matrix}\right.\)\(\)